Bài 1: So sánh
$\frac{2016.2017-1}{2016.2017}$ và $\frac{2017.2018-1}{2017.2018}$ (Hướng dẫn: so sánh phần bù $\frac{1}{2016.2017}$ và $\frac{1}{2017.2018}$)
Bài 2: So sánh
$\frac{2016.2017+1}{2016.2017}$ và $\frac{2017.2018+1}{2017.2018}$ (Hướng dẫn: so sánh phần thừa $\frac{1}{2016.2017}$ và $\frac{1}{2017.2018}$)
Bài 3: So sánh
A = $\frac{2016}{2017}$ + $\frac{2017}{2018}$ và B = $\frac{2016+2017}{2017+2018}$
(Hướng dẫn: Tách B = $\frac{2016}{2017+2018}$ + $\frac{2017}{2017+2018}$)
Bài 4; So sánh
A = $\frac{17^{18}+1}{17^{19}+1}$ và B = $\frac{17^{17}+1}{17^{18}+1}$
Đáp án:
$\begin{array}{l}
B1)\\
A = \dfrac{{2016.2017 – 1}}{{2016.2017}}\\
= \dfrac{{2016.2017}}{{2016.2017}} – \dfrac{1}{{2016.2017}}\\
= 1 – \dfrac{1}{{2016.2017}}\\
B = \dfrac{{2017.2018 – 1}}{{2017.2018}}\\
= 1 – \dfrac{1}{{2017.2018}}\\
Do:2016.2017 < 2017.2018\\
\Rightarrow \dfrac{1}{{2016.2017}} > \dfrac{1}{{2017.2018}}\\
\Rightarrow – \dfrac{1}{{2016.2017}} < – \dfrac{1}{{2017.2018}}\\
\Rightarrow 1 – \dfrac{1}{{2016.2017}} < 1 – \dfrac{1}{{2017.2018}}\\
\Rightarrow A < B\\
B2)A = \dfrac{{2016.2017 + 1}}{{2016.2017}}\\
= \dfrac{{2016.2017}}{{2016.2017}} + \dfrac{1}{{2016.2017}}\\
= 1 + \dfrac{1}{{2016.2017}}\\
B = \dfrac{{2017.2018 + 1}}{{2017.2018}}\\
= 1 + \dfrac{1}{{2017.2018}}\\
Do:2016.2017 < 2017.2018\\
\Rightarrow \dfrac{1}{{2016.2017}} > \dfrac{1}{{2017.2018}}\\
\Rightarrow 1 + \dfrac{1}{{2016.2017}} > 1 + \dfrac{1}{{2017.2018}}\\
\Rightarrow A > B\\
B3)\\
A = \dfrac{{2016}}{{2017}} + \dfrac{{2017}}{{2018}}\\
B = \dfrac{{2016 + 2017}}{{2017 + 2018}}\\
= \dfrac{{2016}}{{2017 + 2018}} + \dfrac{{2017}}{{2017 + 2018}}\\
Do:\dfrac{{2016}}{{2017}} > \dfrac{{2016}}{{2017 + 2018}}\\
\dfrac{{2017}}{{2018}} > \dfrac{{2017}}{{2017 + 2018}}\\
\Rightarrow \dfrac{{2016}}{{2017}} + \dfrac{{2017}}{{2018}} > \dfrac{{2016}}{{2017 + 2018}} + \dfrac{{2017}}{{2017 + 2018}}\\
\Rightarrow A > B\\
B4)\\
A = \dfrac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < 1\\
\Rightarrow \dfrac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < \dfrac{{{{17}^{18}} + 1 + 16}}{{{{17}^{19}} + 1 + 16}}\\
\Rightarrow A < \dfrac{{{{17}^{18}} + 17}}{{{{17}^{19}} + 17}}\\
\Rightarrow A < \dfrac{{17\left( {{{17}^{17}} + 1} \right)}}{{17\left( {{{17}^{18}} + 1} \right)}}\\
\Rightarrow A < \dfrac{{{{17}^{17}} + 1}}{{{{17}^{18}} + 1}}\\
\Rightarrow A < B
\end{array}$